Beta function

From Wikipedia, the free encyclopedia

  (Redirected from Incomplete beta function)
Jump to: navigation, search
This article is about Euler beta function. For other uses, see Beta function (disambiguation).

In mathematics, the beta function, also called the Euler integral of the first kind, is a special function defined by

\mathrm{\Beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt \!

for \textrm{Re}(x), \textrm{Re}(y) > 0.\,

The beta function was studied by Euler and Legendre and was given its name by Jacques Binet.

Contents

[edit] Properties

The beta function is symmetric, meaning that

\Beta(x,y) = \Beta(y,x). \!

It has many other forms, including:

\Beta(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \!
\Beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \textrm{Re}(x)>0,\ \textrm{Re}(y)>0 \!
\Beta(x,y) = \int_0^\infty\dfrac{t^{x-1}}{(1+t)^{x+y}}\,dt, \qquad \textrm{Re}(x)>0,\ \textrm{Re}(y)>0 \!
\Beta(x,y) = \sum_{n=0}^\infty \dfrac{{n-y \choose n}} {x+n}, \!
\Beta(x,y) = \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1}, \!
\Beta(x,y) \cdot \Beta(x+y,1-y) = \dfrac{\pi}{x \sin(\pi y)}, \!


where \Gamma\, is the gamma function. The second identity shows in particular \Gamma(1/2) = \sqrt \pi. Some of these identities, e.g. the trigonometric formula, can be applied to deriving the volume of an n-ball in Cartesian coordinates.

Just as the gamma function for integers describes factorials, the beta function can define a binomial coefficient after adjusting indices:

{n \choose k} = \frac1{(n+2) \Beta(n-k+1, k+1)}.

Moreover, for integer n, \Beta\, can be integrated to give a closed form, an interpolation function for continuous values of k:

{n \choose k} = (-1)^n n! \cfrac{\sin (\pi k)}{\pi \prod_{i=0}^n (k-i)}.

The beta function was the first known scattering amplitude in string theory, first conjectured by Gabriele Veneziano. It also occurs in the theory of the preferential attachment process, a type of stochastic urn process.

[edit] Relationship between gamma function and beta function

To derive the integral representation of the beta function, write the product of two factorials as

\Gamma(x)\Gamma(y) = \int_0^\infty\ e^{-u} u^{x-1}\,du \int_0^\infty\ e^{-v} v^{y-1}\,dv. \!

Now, let u \equiv a^2, v \equiv b^2, so

\begin{align} \Gamma(x)\Gamma(y) & {} = 4\int_0^\infty\ e^{-a^2} a^{2x-1}\mathrm{d}a \int_0^\infty\ e^{-b^2} b^{2y-1}\,db \\ & {} = \int_{-\infty}^\infty\ \int_{-\infty}^\infty\ e^{-(a^2+b^2)} |a|^{2x-1} |b|^{2y-1} \,da \,db. \end{align} \!

Transforming to polar coordinates with a = rcosθ, b = rsinθ:

\begin{align} \Gamma(x)\Gamma(y) & {} = \int_0^{2\pi}\ \int_0^\infty\ e^{-r^2} |r\cos\theta|^{2x-1} |r\sin\theta|^{2y-1} r \, dr \,d\theta \\ & {} = \int_0^\infty\ e^{-r^2} r^{2x+2y-2} r\, dr \int_0^{2\pi}\ |(\cos\theta)^{2x-1} (\sin\theta)^{2y-1}| \, d\theta \\ & {} = \frac{1}{2}\int_0^\infty\ e^{-r^2} r^{2(x+y-1)} \, d(r^2) 4\int_0^{\pi/2}\ (\cos\theta)^{2x-1} (\sin\theta)^{2y-1} \,d\theta \\ & {} = \Gamma(x+y) 2\int_0^{\pi/2}\ (\cos\theta)^{2x-1} (\sin\theta)^{2y-1} \, d\theta \\ & {} = \Gamma(x+y) \Beta(x,y). \end{align}

Hence, rewrite the arguments with the usual form of beta function:

\Beta(x,y) = \frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}.

For another derivation, notice that the stated identity also follows as a particular case of the identity for the integral of a convolution. Taking

f(u):=e^{-u} u^{x-1} 1_{\R_+} and g(u):=e^{-u} u^{y-1} 1_{\R_+}, it is easy to check that:
\Gamma(x)\Gamma(y)=\left(\int_{\R}f(u)du\right)\left(\int_{\R}g(u)du\right)=\int_{\R}(f*g)(u)du=\Beta(x, y)\,\Gamma(x+y).

[edit] Derivatives

The derivatives follow:

{\partial \over \partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \left( {\Gamma'(x) \over \Gamma(x)} - {\Gamma'(x + y) \over \Gamma(x + y)} \right) = \mathrm{B}(x, y) (\psi(x) - \psi(x + y))

where \ \psi(x) is the digamma function.

[edit] Integrals

The Nörlund-Rice integral is a contour integral involving the beta function.

[edit] Approximation

Stirling's approximation gives the asymptotic formula

\Beta(x,y) \sim \sqrt {2\pi } \frac{{x^{x - \frac{1}{2}} y^{y - \frac{1}{2}} }}{{\left( {x + y} \right)^{x + y - \frac{1}{2}} }}

for large x and large y. If on the other hand x is large and y is fixed, then

\Beta(x,y) \sim \Gamma(y)\,x^{-y}.

[edit] Incomplete beta function

The incomplete beta function, a generalization of the beta function, is defined as

\Beta(x;\,a,b) = \int_0^x t^{a-1}\,(1-t)^{b-1}\,dt. \!

For x = 1, the incomplete beta function coincides with the complete beta function. The relationship between the two functions is like that between the gamma function and its generalization the incomplete gamma function.

The regularized incomplete beta function (or regularized beta function for short) is defined in terms of the incomplete beta function and the complete beta function:

I_x(a,b) = \dfrac{\Beta(x;\,a,b)}{\Beta(a,b)}. \!

Working out the integral (one can use integration by parts to do that) for integer values of a and b, one finds:

I_x(a,b) = \sum_{j=a}^{a+b-1} {(a+b-1)! \over j!(a+b-1-j)!} x^j (1-x)^{a+b-1-j}.

The regularized incomplete beta function can be used to evaluate the cumulative density function of a random variable X from a binomial distribution, where the "probability of success" is p and the sample size is n:

F(k;n,p) = \Pr(X \le k) = I_{1-p}(n-k, k+1).

[edit] Properties

I_0(a,b) = 0 \,
I_1(a,b) = 1 \,
I_x(a,b) = 1 - I_{1-x}(b,a) \,

(Many other properties could be listed here.)

[edit] See also

[edit] References

[edit] External links

Retrieved from "http://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function"